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Permo Combo continues 6

by maihuna » Sun Aug 09, 2009 10:12 am
Five straight lines are drawn in one place so that each line cuts all the others and all such point of intersections are distinct points. How many triangles can be formed from such intersections.

5
10
15
20
30
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by DanaJ » Sun Aug 09, 2009 12:18 pm
Pfff... Weird one!

I tried to figure out the number of intersection points "experimentally": two lines intersect in one point, three lines intersect in three points, four lines intersect in 6 points. It's obvious to me that the number of intersection points will be:

1 + 2 + 3 + ... + (n-1), where n is the total number of lines.

So you get intersection points = 1 + 2 + 3 + 4 = 10. Unfortunately, here's where my brain stopped working and couldn't figure out a solution, so I immediately drew the thing and counted 10 triangles. Why it's so (if I'm right) or what's the formula, beats me.
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by real2008 » Sun Aug 09, 2009 12:19 pm
it will be 10, i suppose
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by ankitns » Sun Aug 09, 2009 12:36 pm
Interesting...

I agree that there will be 10 intersection points. This is similar to questions like..."how many hand shakes in the room" or "how many games are played between X teams if each one plays the other once"...etc...

so the formula is essentially sum of all the integers from 1 to (n-1). In this case we have 4+3+2+1 = 10.


Now for the second half of the question. How many triangles, this translates to how many different ways are there to choose 3 points out of the 10 points. Since each unique combination of 3 points leads to a different triangle. But 10choose3 gives us 120. This is not one of the answers so dont know..

It could be that i just have the wrong approach...but would like to know what everyone else thinks.

What is the source of this question? How reliable is it?


Thanks.
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by prindaroy » Sun Aug 09, 2009 7:12 pm
The answer is 10. 5c3 = 10

5 lines, to form a triangle, you need 3 lines, so how many ways can you choose 3 out 5?

5c3 = 10
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by Naruto » Mon Aug 10, 2009 3:21 am
prindaroy wrote:The answer is 10. 5c3 = 10

5 lines, to form a triangle, you need 3 lines, so how many ways can you choose 3 out 5?

5c3 = 10
I disagree with your logic, the question is how many triangles can be formed with all the available points of intersection, not how many arrangements are possible to form one triangle. Your solution is for the latter.
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by prindaroy » Mon Aug 10, 2009 8:03 am
With 3 lines, you can form 3 distinct points of intersections. With four lines you can form four distinct points of intersection. So for 5, you can point five distinct points of intersection. So think about it this way. Let's say with five points of intersection, you have a pentagon. How many triangles can you make out of a pentagon? You can make 10 triangles because 5c3. I know my logic is correct because this is a kaplan question which I answered a few days ago.
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by DanaJ » Mon Aug 10, 2009 12:07 pm
prindaroy wrote:With 3 lines, you can form 3 distinct points of intersections. With four lines you can form four distinct points of intersection. So for 5, you can point five distinct points of intersection. So think about it this way. Let's say with five points of intersection, you have a pentagon. How many triangles can you make out of a pentagon? You can make 10 triangles because 5c3. I know my logic is correct because this is a kaplan question which I answered a few days ago.
Actually, with four lines you will have 6 different points of intersection and with 5 lines you'll have 10. Draw the whole thing to double check. This is why your reasoning is flawed.
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by tohellandback » Tue Aug 11, 2009 12:13 am
I think the solution 5C3 is correct to get the number of triangles. It works for all the numbers.
the reasoning behind it: if someone can post..
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by ankitns » Mon Aug 24, 2009 7:16 pm
5C3 is correct but the logic behind it is a little fuzzy. I just reread the question and it seems that some of us misread "formed from such intersections" as "formed from such intersection points"...

since we do not need to consider the points..we only need to choose 3 different lines...3 non-parallel lines form a unique triangle...the total ways of choosing 3 lines from 5 is 5C3...hence the answer is 5C3..

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prindaroy wrote:With 3 lines, you can form 3 distinct points of intersections. With four lines you can form four distinct points of intersection. So for 5, you can point five distinct points of intersection. So think about it this way. Let's say with five points of intersection, you have a pentagon. How many triangles can you make out of a pentagon? You can make 10 triangles because 5c3. I know my logic is correct because this is a kaplan question which I answered a few days ago.
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